I have some answers for these but do not know how to porperly upload them. I wou
ID: 1060532 • Letter: I
Question
I have some answers for these but do not know how to porperly upload them. I would appreciate it if you could answer these questons and explain. I am unclear on this section of o chem since it wasn't taught in my lecture. thank you
7. Which structure is most consistent with the IR spectrum shown below? (2 pts) 2000 1000 1500 WAVENUMBER I-11 OH OH 8. If we wished to monitor a frequency to determine whether the reaction below had taken place, which frequency would be the best choice? (2 pts) CH3I a. 2200 cm d. 1050 cm 1 b. 1600 cm e. 2950 cm c. 3300 cm 1Explanation / Answer
7) The peak near 1680-1750 cm^-1 is due to carbonyl stretching and this is one of the strongest IR absorptions
so the compound must have a carbonyl group
The broadened peak above 3000 cm^-1 can be attributed to OH group in the compound
so the compound must be a carboxylic acid [as given it will be propanoic acid, structure b.
8) The formation of tertiary amine from secondary amine results in loss of N-H stretching which appears near 3300-3500 cm^-1 hence the peak will be 3300
5) There will be no quartet as there is no hydrogen with three hydrogens on nearby carbon.
6) Ha: there are two hydrogens on nearby carbon so it will split as triplet
Hb:there are two hydrogens on nearby carbon so it will split as triplet
HC: there are two hydrogens on nearby carbon so it will split as triplet
HD: there are two hydrogens on nearby carbon so it will split as triplet
2) the degree of unsaturation is calculated as
Degree of unsaturation = [C +1] - [ H - N + X]/2
C= number of carbon atoms
H = Number of hydrogen atoms
N = Number of nitrogen atoms
X = number of halogen atoms
C5H8Br2
DOU = [5+1] - [ 8 +2]/2 = 6-5 =1
3) singlet means no hydrogen on near by carbon of the given hydrogen
The peak near 2.1 is due to O=C-CH2- hydrogen
So answer is II
4) smallest integration value means that the number of hydrogens of this kind will be least
Answer: HA